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Answer to the Friday Puzzle…..

Imagine that there are 2 trees in a garden. Let us call them Tree A and Tree B. Now imagine that there some birds on both trees.

The birds on Tree A say to the birds on Tree B: ‘If one of you comes to our tree, then our population will be the double of yours.’

Then birds on Tree B say to the birds on Tree A: ‘If one of you comes to our tree, then our population will be equal to that of yours.’

How many birds are there in each tree?

If you have not tried to solve it, have a go now. For everyone else the answer is after the break.

There are 7 birds on Tree A, and 5 on Tree B. Did you solve it?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle(UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.

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39 comments on “Answer to the Friday Puzzle…..”

Tried and failed to solve this with algebra. Must be rusty. But I solved it thus:

Assume each tree contains whole birds.
If A+1 is *double* B-1 then A must be an even number.
If B+1 is the same as A-1 then A must be B+2.
B must be an odd number that is two less than A.
Given that A+1 = 2(B-1) and B+1=A-1 then A and B must be small numbers.
Therefore, candidates are 3,1;5,3; 7,5; 9,7.
It is trivial to determine which of these four pairs is the correct answer
A=7; B=5.

I know what you mean by rusty, lol, but to shake off the rust, you just have to read slowly and rewrite what the problem says using letters(variables) This gives you TreeA=X TreeB=Y …. Taking 1 from B and adding it to A, makes treeA twice that of treeB so… X+1=2(Y-1) And taking 1 from treeA and giving it to B makes them equal so …. X-1=Y+1 So simplifying gives X-2Y+3=0 and X-Y-2=0 and you just have to solve these simultaneous equations! Cheers!

Anon- I got 6 and 4 with algebra. It was an error with the brackets. It’s along the lines of 2(b-1) which = 2b-2 not 2b-1 which is where I went wrong.
I got the answer in the end by working out that the answer had to add up to an even number and be divisible by 3 . Then I got my penny jar out to work out the combinations .

This is what I did, but it’s been so long since I took algebra, is that the “simultaneous equations” approach people are talking about? I remember learning to do simultaneous equations, but nothing concrete about how. I just figured if I could isolate one of the variables in both equations, then I could set the other sides equal to each other and solve for one variable. Did I remember the correct approach, or just happen to make up something that also worked?

I found a 2nd solution, (3 on each tree) because of my interpretation of the word “then”.
“Then birds on Tree B say to the birds on Tree A: ‘If one of you comes to our tree, then our population will be equal to that of yours.’ ”
I presumed you were being tricky and I interpreted the word “then” as meaning AFTER one of the B birds flew to A.
Stan

That doesn’t work though, because the words spoken by the birds start “If”; the numbers are all hypothetical: nowhere does it say that any of the birds actually fly between the trees.

The “then”, however, is said by the narrator, not by any of the birds. So the “then” doesn’t feature in the hypothetical bird-moving. If the second set of birds to speak had started their statement with “then”, that would’ve been the birds saying their statement follows on from the situation established by the first birds.

But because the then is merely describing the order in which the birds speak, both statements made are hypothetical alternatives based on the current situation.

The quick and easy way is thus. Clearly answers must be fairly small numbers. Since in first move, there are twice as many in one tree as in other, the total number must be divisible by 3. Since in second move there are equal birds in each tree, total must be even. So number of birds is divisible by 6. 6 clearly isn’t enough. Try 12, oh yes that words.

While its true that this is such a simple problem that using algebra may not be necessary, its still a good idea to understand how to do it using algebraic equations because if the problem was even slightly more complicated you would not be able to “reason” it out. Knowing how to extract the equations becomes integral to solving harder problems, so its good practice.

Again, there is no “puzzle” here, especially for the intelligent people who read this blog, its just straight algebra! I guess though it is good to practice your algebra skills, you never know when it may come in handy. The equations were: X-2Y+3=0 and X-Y-2=0 giving X=7(treeA) and Y=5(treeB)